Exercises, examples & solutions - Classical Mechanics & Special Relativity for Starters

According to $S'$

\begin{split} U'_0 &= \gamma(U')c = \frac{41}{9} c \\ U'_1 &= \gamma(U') U' = \frac{40}{9} c \end{split}

(3)

LT to $S$ using $\gamma (V) = \frac{13}{5}$ :

\begin{split} U_0 &= \gamma(V) \left ( U'_0 + \frac{V}{c} U'_1) \right ) = \frac{13}{5} \left ( \frac{41}{9} c + \frac{12}{13} \frac{40}{9} c \right ) = \frac{1013}{45}c\\ U_1 &= \gamma(V) \left (U'_1 + \frac{V}{c} U'_0) \right ) = = \frac{13}{5} \left ( \frac{40}{9} c + \frac{12}{13} \frac{41}{9} c \right ) = \frac{1012}{45}c \end{split}

(4)

We find $u_x$ by taking the ratio $\frac{U_1}{U_0} = \frac{\gamma(U)u}{\gamma(U)c}$ :

\begin{split} u_x &= \frac{1012}{1013} c \lt 1\\ u_y &= u_z =0 \end{split}

(5)

Solution to Exercise 2

According to $S'$

\begin{split} U'_0 &= \gamma(U')c = \frac{41}{9} c \\ U'_1 &= 0 \\ U'_2 &= \gamma(U') U' = \frac{40}{9} c \end{split}

(6)

LT naar $S$ using $\gamma (V) = \frac{13}{5}$ :

\begin{split} U_0 &= \gamma(V) \left ( U'_0 + \frac{V}{c} U'_1) \right ) = \frac{13}{5} \left ( \frac{41}{9} c + 0 \right ) = \frac{533}{45}c\\ U_1 &= \gamma(V) \left (U'_1 + \frac{V}{c} U'_0) \right ) = = \frac{13}{5} \left ( 0 + \frac{12}{13} \frac{41}{9} c \right ) = \frac{492}{45}c \\ U_2 &= U'_2 = \frac{40}{9} c \end{split}

(7)

We find $u_x$ by taking the ratio $\frac{U_1}{U_0} = \frac{\gamma(U)u_x}{\gamma(U)c}$ :

u_x = \frac{492}{533} c

(8)

Similarly:

u_y = \frac{U_2}{U_0} = \frac{\gamma(U)u_y}{\gamma(U)c} = \frac{40}{533}c

(9)

The magnitude of the velocity according to $S4$ is

u =\sqrt{u^2_x + u^2_y} = \sqrt{\frac{243664}{284089}} c \approx 0.93 c \lt 1 c

(10)

Solution to Exercise 3

According to $S'$ the photon is send at $E_1: (ct'_1, x'_1 ) = (0, 1) ls$ . Thus, it is received at $E_2: (ct'_2, x'_2 ) = (1,0)$ . Hence, for $S$ event $E_1$ has coordinates:

\begin{split} ct_1 &= \frac{5}{4} \left ( 0 + \frac{3}{5} 1 \right ) = \frac{3}{4} ls \\ x_1 &= \frac{5}{4} \left ( 1 + \frac{3}{5} 0 \right ) = \frac{5}{4} ls \end{split}

(11)

and thus, $S$ receives this photon at $(ct_3, x_3) = ( 2, 0)ls$ .

For $S'$ the 4-Momentum of the photon is: $\left ( \frac{hf_0}{c}, -\frac{hf_0}{c}\right )$ . If we transform this to the frame of $S$ , we get:

\frac{hf}{c} = \frac{5}{4} \left ( \frac{hf_0}{c} + \frac{3}{5} \cdot -\frac{hf_0}{c} \right ) = \frac{1}{2}\frac{hf_0}{c} \Rightarrow f =\frac{1}{2}f_0

(12)

Solution to Exercise 4

In this case for $S'$ the 4-momentum of the photon is:

P'^\mu = \left ( \frac{hf_0}{c},0, \pm \frac{hf_0}{c}, 0 \right )

(13)

If we translate this to the world of $S$ , we need to realize that momentum is a vector and that the spatial parts, i.e. $P^1, P^2, P^3$ form a 3-vector. In this case, there is no $z$ -component and we can write the $x$ and $y$ -components as the length of the vector times a $\cos$ and a $\sin$ , respectively:

\begin{split} \frac{hf}{c} &= \frac{5}{4} \left (\frac{hf_0}{c} + \frac{3}{5} 0 \right ) = \frac{5}{4} \frac{hf_0}{c}\\ \frac{hf}{c} \cos \alpha & = \frac{5}{4} \left ( 0 + \frac{3}{5}\frac{hf_0}{c}\right ) = \frac{3}{4} \frac{hf_0}{c}\\ \frac{hf}{c} \sin \alpha & = \pm \frac{hf_0}{c} \end{split}

(14)

Thus, from the time-like component we conclude: $f = \frac{5}{4}f_0$ . This should be in agreement with the spatial components. Let’s check:

\begin{split} \frac{h^2f^2}{c^2} &= \frac{h^2f^2}{c^2} \cos^2 \alpha + \frac{h^2f^2}{c^2} \sin^2 \alpha \\ &= \frac{3^2}{4^2} \frac{h^2f_0^2}{c^2} + \frac{h^2f_0^2}{c^2} \\ &= \frac{5^2}{4^2} \frac{h^2f_0^2}{c^2} \end{split}

(15)

Indeed, the two spatial components are in agreement with the time-like one.

Finally, we have that according to $S$ , the photon travels at an angle $\tan \alpha = \pm \frac{4}{3} \rightarrow \alpha = \pm 53.13^\circ$ with the $x$ -axis.

5.7.1.1Exercises¶

5.7.1.2Answers¶

5.7.1Exercises, examples & solutions

5.7.1.1Exercises¶

5.7.1.2Answers¶